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XTTX
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OK, well this is probably a pretty basic problem. I understand all of it except for one part.
If F(x)=\int^{x}_{0} sin(t)dt, where x \geq 0, what is the maximum value of F?
F(x)=\int^{x}_{0} sin(t)dt
F(x)=\int^{x}_{0} sin(t)dt = -cos(x) + c
max = 1 + c
How do you find c?
If solving with a graphing calculator: y^{}_{1} = Fnint(sin(t),t,0,x) then the answer is 2; I just don't know how the calculator found the shift.
Homework Statement
If F(x)=\int^{x}_{0} sin(t)dt, where x \geq 0, what is the maximum value of F?
Homework Equations
F(x)=\int^{x}_{0} sin(t)dt
The Attempt at a Solution
F(x)=\int^{x}_{0} sin(t)dt = -cos(x) + c
max = 1 + c
How do you find c?
If solving with a graphing calculator: y^{}_{1} = Fnint(sin(t),t,0,x) then the answer is 2; I just don't know how the calculator found the shift.
That is, your solution F(x) is actually [-cos(x) + C] - [-cos(0) + C], and the C cancels out.